For $c0,$ if $a\hat{i}+b\hat{j}+c\hat{k}$ is the unit normal vector at $(1,1,\sqrt{2})$ to the cone $z=\sqrt{x^{2}+y^{2}}$, then :

Question

For $c>0,$ if $a\hat{i}+b\hat{j}+c\hat{k}$ is the unit normal vector at $(1,1,\sqrt{2})$ to the cone $z=\sqrt{x^{2}+y^{2}}$, then :

Accepted Answer

To determine why Option 2 is the correct answer, we need to analyze the problem and the given options.

The problem involves finding the unit normal vector to the cone defined by the equation $ z = \sqrt{x^2 + y^2} $ at the point $ (1, 1, \sqrt{2}) $.

### Step-by-step Explanation:

1. **Equation of the Cone:**
   The cone is given by $ z = \sqrt{x^2 + y^2} $. This can be rewritten as $ z^2 = x^2 + y^2 $.

2. **Gradient Vector:**
   To find the normal vector to the surface at a given point, we calculate the gradient of the function $ f(x, y, z) = z^2 - x^2 - y^2 $. The gradient is given by:
   $$
   \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (-2x, -2y, 2z)
   $$

3. **Evaluate at the Point (1, 1, $\sqrt{2}$):**
   Substituting $ x = 1 $, $ y = 1 $, and $ z = \sqrt{2} $ into the gradient, we get:
   $$
   \nabla f(1, 1, \sqrt{2}) = (-2 \times 1, -2 \times 1, 2 \times \sqrt{2}) = (-2, -2, 2\sqrt{2})
   $$

4. **Unit Normal Vector:**
   The unit normal vector is obtained by normalizing the gradient vector:
   $$
   \text{Magnitude of } \nabla f = \sqrt{(-2)^2 + (-2)^2 + (2\sqrt{2})^2} = \sqrt{4 + 4 + 8} = \sqrt{16} = 4
   $$
   Thus, the unit normal vector is:
   $$
   \left( \frac{-2}{4}, \frac{-2}{4}, \frac{2\sqrt{2}}{4} \right) = \left( -\frac{1}{2}, -\frac{1}{2}, \frac{\sqrt{2}}{2} \right)
   $$

5. **Verification with Options:**
   For the unit normal vector $ a\hat{i} + b\hat{j} + c\hat{k} = -\frac{1}{2}\hat{i} -\frac{1}{2}\hat{j} + \frac{\sqrt{2}}{2}\hat{k} $, we have $ a = -\frac{1}{2} $, $ b = -\frac{1}{2} $, and $ c = \frac{\sqrt{2}}{2} $.

Now, calculate $ a^2 + b^2 - c^2 $:
   $$
   a^2 + b^2 - c^2 = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 - \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} - \frac{2}{4} = \frac{2}{4} - \frac{2}{4} = 0
   $$

This confirms that Option 2 is correct.

### Why Other Options are Incorrect:

- **Option 1: $ a^2 + b^2 + c^2 = 0 $**
  - This would imply that the vector is the zero vector, which is not possible for a unit normal vector.

- **Option 3: $ a^2 - b^2 + c^2 = 0 $**
  - Substituting the values, we get $ \frac{1}{4} - \frac{1}{4} + \frac{2}{4} = \frac{2}{4} \neq 0 $.

- **Option 4: $ -a^2 + b^2 + c^2 = 0 $**
  - Substituting the values, we get $ -\frac{1}{4} + \frac{1}{4} + \frac{2}{4} = \frac{2}{4} \neq 0 $.

Thus, Option 2 is the only correct choice, as it satisfies the condition for the unit normal vector at the given point on the cone.